Let $R$ be a ring with unity. Let $A$ and $B$ be any right $R$-modules. Recall that $A$ is said to be $B$-injective if every homomorphism $f:B'\to A$, where $B'$ is a submodule of $B$, can be extended to a homomorphism $g:B\to A$. We say that $A$ and $B$ are relatively injective if $A$ is $B$-injective and $B$ is $A$-injective.
Now, assume that $Hom_R(A,B)=0$ and $Hom_R(B,A)=0$. Does this imply that $A$ and $B$ are relatively injective?. If not, what can we conclude about the modules $A$ and $B$ is such conditions hold?. I claim that $A$ and $B$ must be simples. Is this true?.
Thanks in advance.
The modules $A$ and $B$ need not be relatively injective. Let $K$ be a field, and take $$ R = \begin{bmatrix} K & 0 & K \\ 0 & K & K \\ 0 & 0 & K \end{bmatrix}. $$ Consider the right $R$-modules $A = \begin{bmatrix} K & 0 & K \end{bmatrix}$ and $B = \begin{bmatrix} 0 & K & K \end{bmatrix}$. Then $Hom_R(A,B) = 0 = Hom_R(B,A)$, but they are not relatively injective, since they have a common non-zero submodule $C = \begin{bmatrix} 0 & 0 & K \end{bmatrix}$.
If you are used to the language of quiver representations, the algebra $R$ above is the path algebra of the quiver $1\rightarrow 3 \leftarrow 2$ and the modules $A$ and $B$ are the represenations $K\xrightarrow{1} K \xleftarrow{} 0$ and $0 \xrightarrow{} K \xleftarrow{1} K$, respectively.